function logout()
{
$session_ida = $_SESSION['user']['name'];
$query3 = "UPDATE 'userlog' set 'logout' = CURTIME() where username =
'$session_ida'";
mysqli_query($db,$query3);
unset($_SESSION['user']['name']);
session_destroy();
header("location: login.php");
}
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Devon Bessemer
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1$db doesn't exist in this scope. Enable error reporting and always include those errors in your questions. – Devon Bessemer Aug 07 '18 at 17:09
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Pass $db as an element to the function call . – PHP Web Aug 07 '18 at 18:17
1 Answers
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You haven't defined the variable $db yet. I'll bet that you defined it in a different scope - however, as this is a separate function, this variable isn't directly accessible.
Joel Rummel
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